contestada

A) An electron is moving east in a uniform electric field of 1.48 N/C directed to the west. At point A, the velocity of the electron is
4.55
×
10
5
m/s pointed towards the east.

What is the speed of the electron when it reaches point B which is a distance of 0.370 m east of point A?

B) A proton is moving in the uniform electric field of part (A), at which point A, the velocity of the proton is
1.91
×
10
4
m/s, again points towards the east.

What is the speed of the proton at point B?

Respuesta :

olatunjiolorundare20

Answer:A)3074m,3074.37m/s

B)12905.4m,19100.54m/s

Explanation:the formula for electric field =V/d

Where V=velocity of the particle,d=distance covered from the electric field source

A)E=V/d

1.48=4.55×10^5/d

d=4.55×10^5/1.48=3074m

Distance of B from centre of the field=3074+0.370=3074.370m

B)d=V/E=1.91×10^4/1.48

d=12905.4m

At point B

Distance=12905.4+0.370=12905.770

V=Ed=12905.770×1.48=19100.54m/s

Otras preguntas